Integral Test

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Theorem

Let $f$ be a real function which is continuous, positive and decreasing on the interval $\hointr 1 {+\infty}$.

Let the sequence $\sequence {\Delta_n}$ be defined as:

$\displaystyle \Delta_n = \sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x$


Then $\sequence {\Delta_n} $ is decreasing and bounded below by zero.

Hence it converges.


Proof

From Upper and Lower Bounds of Integral, we have that:

$\displaystyle m \paren {b - a} \le \int_a^b \map f x \rd x \le M \paren {b - a}$

where:

$M$ is the maximum

and:

$m$ is the minimum

of $\map f x$ on $\closedint a b$.

Since $f$ decreases, $M = \map f a$ and $m = \map f b$.

Thus it follows that:

$\displaystyle \forall k \in \N_{>0}: \map f {k + 1} \le \int_k^{k + 1} \map f x \rd x \le \map f k$

as $\paren {k + 1} - k = 1$.


Thus:

\(\displaystyle \Delta_{n + 1} - \Delta_n\) \(=\) \(\displaystyle \paren {\sum_{k \mathop = 1}^{n + 1} \map f k - \int_1^{n + 1} \map f x \rd x} - \paren {\sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x}\)
\(\displaystyle \) \(=\) \(\displaystyle \map f {n + 1} - \int_n^{n + 1} \map f x \rd x\)
\(\displaystyle \) \(\le\) \(\displaystyle \map f {n + 1} - \map f {n + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)


Thus $\sequence {\Delta_n}$ is decreasing.

$\Box$


Also:

\(\displaystyle \Delta_n\) \(=\) \(\displaystyle \sum_{k \mathop = 1}^n \map f k - \sum_{k \mathop = 1}^{n - 1} \int_k^{k + 1} \map f x \rd x\)
\(\displaystyle \) \(\ge\) \(\displaystyle \sum_{k \mathop = 1}^n \map f k - \sum_{k \mathop = 1}^{n - 1} \map f k\)
\(\displaystyle \) \(=\) \(\displaystyle \map f n\)
\(\displaystyle \) \(\ge\) \(\displaystyle 0\)

$\Box$


Hence the result.

$\blacksquare$


Also known as

The integral test is also known as the Euler-Maclaurin Summation Formula, but that result properly refer to a more precise theorem of which this is a simple corollary.


Notes

It follows from this that if $f$ is continuous, positive and decreasing on $\hointr 1 \infty$, then the series $\displaystyle \sum_{k \mathop = 1}^\infty \map f k$ and the improper integral $\displaystyle \int_1^{\mathop \to +\infty} \map f x \rd x$ either both converge or both diverge.

So this theorem provides a test for the convergence of both a series and an improper integral.


Sources